Problem: Find the slope and y-intercept of the line that is ${\text{parallel}}$ to $\enspace {y = -\dfrac{1}{3}x + 5}\enspace$ and passes through the point ${(-5, 5)}$. ${1}$ ${2}$ ${3}$ ${4}$ ${5}$ ${6}$ ${7}$ ${8}$ ${9}$ ${\llap{-}2}$ ${\llap{-}3}$ ${\llap{-}4}$ ${\llap{-}5}$ ${\llap{-}6}$ ${\llap{-}7}$ ${\llap{-}8}$ ${\llap{-}9}$ ${1}$ ${2}$ ${3}$ ${4}$ ${5}$ ${6}$ ${7}$ ${8}$ ${9}$ ${\llap{-}2}$ ${\llap{-}3}$ ${\llap{-}4}$ ${\llap{-}5}$ ${\llap{-}6}$ ${\llap{-}7}$ ${\llap{-}8}$ ${\llap{-}9}$
Answer: Parallel lines have the same slope. The slope of the blue line is ${-\dfrac{1}{3}}$ , so the equation of our parallel line will be of the form $\enspace {y = -\dfrac{1}{3}x + b}\enspace$ We can plug our point, $(-5, 5)$ , into this equation to solve for ${b}$ , the y-intercept. $5 = {-\dfrac{1}{3}}(-5) + {b}$ $5 = \dfrac{5}{3} + {b}$ $5 - \dfrac{5}{3} = {b} = \dfrac{10}{3}$ The equation of the parallel line is $\enspace {y = -\dfrac{1}{3}x + \dfrac{10}{3}}\enspace$. ${m = -\dfrac{1}{3}, \enspace b = \dfrac{10}{3}}$